Problem: Simplify and expand the following expression: $ \dfrac{5}{5r - 25}- \dfrac{1}{4r + 32}+ \dfrac{2r}{r^2 + 3r - 40} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $5$ out of denominator in the first term: $ \dfrac{5}{5r - 25} = \dfrac{5}{5(r - 5)}$ We can factor a $4$ out of denominator in the second term: $ \dfrac{1}{4r + 32} = \dfrac{1}{4(r + 8)}$ We can factor the quadratic in the third term: $ \dfrac{2r}{r^2 + 3r - 40} = \dfrac{2r}{(r - 5)(r + 8)}$ Now we have: $ \dfrac{5}{5(r - 5)}- \dfrac{1}{4(r + 8)}+ \dfrac{2r}{(r - 5)(r + 8)} $ The least common multiple of the denominators is: $ 20(r - 5)(r + 8)$ In order to get the first term over $20(r - 5)(r + 8)$ , multiply by $\dfrac{4(r + 8)}{4(r + 8)}$ $ \dfrac{5}{5(r - 5)} \times \dfrac{4(r + 8)}{4(r + 8)} = \dfrac{20(r + 8)}{20(r - 5)(r + 8)} $ In order to get the second term over $20(r - 5)(r + 8)$ , multiply by $\dfrac{5(r - 5)}{5(r - 5)}$ $ \dfrac{1}{4(r + 8)} \times \dfrac{5(r - 5)}{5(r - 5)} = \dfrac{5(r - 5)}{20(r - 5)(r + 8)} $ In order to get the third term over $20(r - 5)(r + 8)$ , multiply by $\dfrac{20}{20}$ $ \dfrac{2r}{(r - 5)(r + 8)} \times \dfrac{20}{20} = \dfrac{40r}{20(r - 5)(r + 8)} $ Now we have: $ \dfrac{20(r + 8)}{20(r - 5)(r + 8)} - \dfrac{5(r - 5)}{20(r - 5)(r + 8)} + \dfrac{40r}{20(r - 5)(r + 8)} $ $ = \dfrac{ 20(r + 8) - 5(r - 5) + 40r} {20(r - 5)(r + 8)} $ Expand: $ = \dfrac{20r + 160 - 5r + 25 + 40r}{20r^2 + 60r - 800} $ $ = \dfrac{55r + 185}{20r^2 + 60r - 800}$ Simplify: $ = \dfrac{11r + 37}{4r^2 + 12r - 160}$